The standard reduction potential at 298 K for the following half cells are given below:- \(\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O} \quad \mathrm{E}^{0}=0.97 \mathrm{~V}\) \(\mathrm{V}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{V}\) \(\mathrm{E}^{0}=-1.19 \mathrm{~V}\) \(\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}\) \(\mathrm{E}^{0}=-0.04 \mathrm{~V}\) \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})\) \(\mathrm{E}^{0}=0.80 \mathrm{~V}\) \(\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s})\) \(\mathrm{E}^{0}=1.40 \mathrm{~V}\) The number of metal(s) which will be oxidized by \(\mathrm{NO}_{3}^{-}\) in aqueous solution is _____.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Redox Reactions and Electrochemistry

The standard reduction potential at 298 K for the following half cells are given below:-

NO_{3}^{-} + 4 H^{+} + 3 e^{-} \to NO (g) + 2 H_{2} O E^{0} = 0.97 V

V^{2 +} (aq) + 2 e^{-} \to V

E^{0} = - 1.19 V

Fe^{3 +} (aq) + 3 e^{-} \to Fe

E^{0} = - 0.04 V

Ag^{+} (aq) + e^{-} \to Ag (s)

E^{0} = 0.80 V

Au^{3 +} (aq) + 3 e^{-} \to Au (s)

E^{0} = 1.40 V

The number of metal(s) which will be oxidized by

NO_{3}^{-}

in aqueous solution is _____.

Answer: 3

Nitrate acts as the oxidising agent through its half-reaction

N O_{3}^{-} + 4 H^{+} + 3 e^{-} \to NO + 2 H_{2} O

, for which

E^{\circ} = 0.97 V

. A metal is oxidised by nitrate only when its own standard reduction potential is below

0.97 V

, so that the combined cell potential is positive. Vanadium (

- 1.19 V

), iron (

- 0.04 V

) and silver (

0.80 V

) all lie below this value and are oxidised, whereas gold (

1.40 V

) lies above it and is not. The number of metals oxidised by

N O_{3}^{-}

is 3.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer